4 \includegraphics[max width=\textwidth, alt={}, center]{ae809dfc-c5af-4c0a-9c88-009949d3e9f9-3_727_565_1256_790} A uniform lamina of weight 15 N is in the form of a trapezium \(A B C D\) with dimensions as shown in the diagram. The lamina is freely hinged at \(A\) to a fixed point. One end of a light inextensible string is attached to the lamina at \(B\). The lamina is in equilibrium with \(A B\) horizontal; the string is taut and in the same vertical plane as the lamina, and makes an angle of \(30 ^ { \circ }\) upwards from the horizontal (see diagram). Find the tension in the string.

Weight split is 9N:6N

For lamina $9 \times 0.75 + 6 \times 0.5$
$T \times 1.5\sin 30°$
Tension is 13N | $B1, M1, A1 \text{ft}, A1, A1$ [5] | For taking moments about A

**Alternatively**

$[(1.5^2 + 1.5 \times 2)\bar{x} = 1.5^2 \times 0.75 + \frac{1}{2} \times 1.5 \times 2 \times 0.5]$
$\bar{x} = 0.65$ | $M1, A1, M1$ | For using $A\bar{x} = A_1x_1 + A_2x_2$. For taking moments about A

$15 \times 0.65 = T \times 1.5\sin 30°$
Tension is 13N | $A1 \text{ft}, A1$ [5] |

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\includegraphics[max width=\textwidth, alt={}, center]{ae809dfc-c5af-4c0a-9c88-009949d3e9f9-3_727_565_1256_790}

A uniform lamina of weight 15 N is in the form of a trapezium $A B C D$ with dimensions as shown in the diagram. The lamina is freely hinged at $A$ to a fixed point. One end of a light inextensible string is attached to the lamina at $B$. The lamina is in equilibrium with $A B$ horizontal; the string is taut and in the same vertical plane as the lamina, and makes an angle of $30 ^ { \circ }$ upwards from the horizontal (see diagram). Find the tension in the string.

\hfill \mbox{\textit{CAIE M2 2010 Q4 [5]}}